Final answer:
The infection rate of exposed phages can be calculated by understanding the log reduction in PFU/mL and considering the dilution factors in the experiment. A two-log reduction signifies a 100-fold decrease in the concentration of infectious particles. Additional detailed experimental data would be needed to accurately determine the infection rate under the influence of the chemicals.
Step-by-step explanation:
To calculate the infection rate of the exposed phages, one would need to understand the reduction in plaque-forming units (PFU) due to the exposure to chemicals. A two-log reduction means that the number of infectious particles is 100 times less in the exposed group than in the control. If the control had an infection rate of 1 and a concentration of 1000 PFU/mL, and the exposed group had just 10 PFU/mL, we could deduce that the exposure to the chemical greatly reduced the infectivity of the phages.
When determining the PFU/mL, we must consider the dilution factor involved in the experiment. For instance, using serial dilutions is a common technique, where a sample is repeatedly diluted to ultimately quantify the concentration of viruses in the original sample. If you observe 500 plaques at the fourth dilution in a ten-fold serial dilution, it implies that the original undiluted lysate would theoretically have 500 x 10^4 PFU/mL.
However, to precisely determine the infection rate of the exposed phages, more detailed experimental data are required, including the time frame of infection, the efficiency of the phages at different concentrations, and the condition of the host cells. Controls and detailed calculations would allow us to determine the impact of the chemicals on the phage's infectivity.